Computing the class group of a quadratic function field

I am asking for a reference in which I can find tools to answer questions like the following: Let $$K=\mathbb{F}_q(X)$$ be a rational function field over the finite field with $$q$$ elements. Let $$E/K$$ be a quadratic extension. How can I compute the class groups of $$K$$?

I'm not interested in the general case, but in the special case where $$E = K(\sqrt{D})$$, where $$D$$ is an irreducible polynomial of even degree $$\deg D=2g+2$$ and $$g,D$$ may vary (that is, statistical results are also welcomed).

Edit. I've been asked to be more specific in my question: Given a finite field $$\mathbb{F}_q$$ and a prime $$\ell$$ (say $$\ell=3$$ to be concrete), can is there always an hyperelliptic curve $$y^2=D(x)$$ with $$D$$ irreducible of degree $$2g+2$$ such that the class group has a order divisible by $$\ell$$?

• Isn't the class group just the group of $\mathbb{F}_q$-rational points on the jacobian of the smooth projective curve $C$ with function field $E$? Your curve here is a hyperelliptic curve, and I think tmagma can computer the points on the jacobian. Mar 25, 2020 at 15:03
• Thanks for you comment Daniel. It indeed is. But I am trying to have something more conceptual. For example, can I find for every finite field, say of characteristic neq 2, an elliptic curve y^2 = D(x), with D of degree 4 and irreducible such that the number of points is even, or divisible by three? Mar 26, 2020 at 9:03
• I see. Well the question you have written above seems very different to the question in your comments. Can you ask a more specific question which has a well-defined answer? Or do you just want a list of all papers which consider class groups of function fields? Mar 26, 2020 at 15:46
• About the question in the comments: Honda-Tate theory tells you exactly which abelian varieties can exist over any finite field and how to calculate their number of rational points via the Weil conjectures. Now they may not all be jacobians. But in the elliptic curve case they are, so I would guess that this would allow you to answer your question. Mar 26, 2020 at 15:48

The answer to the precise question is yes. See Theorem 1.2 of:

"Homological stability for Hurwitz spaces and the Cohen-Lenstra conjecture over function fields," https://arxiv.org/pdf/0912.0325.pdf.

This implies that a positive proportion of imaginary quadratic extensions of $$\mathbb{F}_q(x)$$ has class number divisible by $$\ell$$, at least for $$q$$ large and $$q \not \equiv 1 \bmod \ell$$.

The corresponding infinitude is easier and was known previously in more generality (see the discussions following Theorem 1.2).

This is more of a too-long comment than an answer.

The specific question in the comments, "can I find for every finite field, say of characteristic neq 2, an elliptic curve $$y^2 = D(x)$$, with $$D$$ of degree 4 and irreducible such that the number of points is even" has a nice positive answer: yes, for any irreducible $$D$$. And the same trick applies to produce hyperelliptic examples in every odd genus.

Let $$g \ge 1$$ be odd, and let $$F$$ be an irreducible polynomial of degree $$2g+2$$ over $$\mathbb{F}_q$$ with $$q$$ odd (I will use $$F$$, rather than $$D$$, for the polynomial, so that I can use $$D$$ for divisors). We define a genus-$$g$$ hyperelliptic curve $$H: y^2 = F(x)$$ over $$\mathbb{F}_q$$. Let $$D_\infty$$ be the (degree 2) divisor at infinity on $$H$$.

The polynomial $$F$$ factors over $$\mathbb{F}_{q^2}$$ as $$F = GG'$$ where $$G$$ and $$G'$$ are irreducible polynomials of degree $$g+1$$ over $$\mathbb{F}_{q^2}$$, conjugate over $$\mathbb{F}_q$$. Let $$\{\alpha_1,\ldots,\alpha_{g+1}\}$$ be the roots of $$G$$ and $$\{\alpha_1',\ldots,\alpha_{g+1}'\}$$ be the roots of $$G'$$. Since $$g$$ is odd, we can define divisors $$D := \sum_{i=1}^{g+1}(\alpha_i,0) - ((g+1)/2)D_\infty$$ and $$D' := \sum_{i=1}^{g+1}(\alpha_i',0) - ((g+1)/2)D_\infty$$ on $$H$$, both of degree $$0$$ and both defined over $$\mathbb{F}_{q^2}$$. Now $$D$$ and $$D'$$ are not principal (they correspond to the ideals $$(G(x),y)$$ and $$G'(x),y)$$, but by construction $$2D = (G)$$ and $$2D' = (G')$$, so $$[D]$$ and $$[D']$$ are 2-torsion divisor classes exchanged by Galois. But in fact they are the same class, because $$D + D' = (y)$$; so $$[D] = [D']$$ is an $$\mathbb{F}_q$$-rational divisor class of order 2, and therefore the divisor class group of $$H$$ has even order.